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Correct option is (B) a subgroup of a group of REAL numbers

For explanation: If we consider the abelian group as a group rational numbers under BINARY OPERATION + then it is an example of a subgroup of a group of real numbers.

The correct choice is (a) Commutativity

The EXPLANATION is: Grupoid has closure property; semigroup has closure and ASSOCIATIVE; monoid has closure, associative and identity property; GROUP has closure, associative, identity and INVERSE; the abelian group has group property and commutative.

Correct choice is (a) singular

Best explanation: A singular element can GENERATE a cyclic group. Every element of a cyclic group is a POWER of some specific element which is known as a GENERATOR ‘g’.

Correct choice is (a) (P, *)

The BEST EXPLANATION: An ALGEBRAIC structure (P,*) is CALLED a semigroup if a*(b*c) = (a*b)*c for all a,b,c belongs to S or the elements follow associative property under “*”. (Matrix,*) and (Set of integers,+) are examples of semigroup.

Correct option is (a) with respect to BINARY operation *

The BEST I can EXPLAIN: A non empty set A is called an algebraic structure w.r.t binary operation “*” if (a*B) belongs to S for all (a*b) belongs to S. Therefore “*” is closure operation on ‘A’.

Right CHOICE is (a) order isomorphism

The EXPLANATION is: We know that very σ-complete BOOLEAN algebra is a Boolean algebra. An isomorphism of Boolean algebras is termed as an order isomorphism. All meets and JOINS present in an order isomorphism domain is preserved. HENCE, a Boolean algebra isomorphism preserves all meets and joins in its domain.

Right CHOICE is (B) ϕ^-1∈P(h)

The explanation is: Let, ϕ and σ both can fix h, then we can have ϕ(σ(h)) = ϕ(h) = h. Hence, ϕ∘σ FIXES h and ϕ∘σ∈P(h). Now, all colorings can be fixed by the identity permutation. So ID∈P(h) and if ϕ(h) = h then ϕ^-1(h) = ϕ^-1(ϕ(h)) = id(h) = h which implies that ϕ^-1∈P(h).

The correct option is (a) groups

Explanation: Suppose, there are two functions f1 and f2 which BELONG to the same EQUIVALENCE CLASS since there exists an invariant PERMUTATION say, π(a permutation that does not change the OBJECT itself, but only its representation), such that: f2*π≡f1. So, invariant permutations can form a group, as the product (composition) of invariant permutations is again an invariant permutation.

Correct choice is (d) 1563150

Explanation: There are m^4 + m^2 + 2m ELEMENTS after PERFORMING all rotations. Dividing this by the NUMBER of transformations 4 produces the desired number of distinct colorings \(\frac{m^4 + m^2 + 2m}{4}\). Hence, the number of distinct colorings with 50 COLORS is 1563150.

The correct answer is (b) Algebraic extension property

The BEST I can EXPLAIN: An ALGEBRA A describes the congruence extension property (CEP) if for every B≤A and θ∈Con(B) there exists a φ∈Con(A) such that θ = φ∩(B×B). A class M of ALGEBRAS has the CEP if every algebra in the class has the CEP. The Congruence Extension Property specifically CHARACTERIZES the distributive lattices among all lattices.

The correct ANSWER is (B) even

For explanation I would say: There are FOUR permutations (2, 5), (2, 8), (2, 4) and (3, 6) and so it is an even permutation.

The correct ANSWER is (b) 54

To elaborate: If k^th ROW of a matrix with n^th row is linearly independent then the RANK of that matrix is k. If we take the transpose of a matrix the rank does not change. Hence, the answer is 54 in this CASE.

Correct ANSWER is (b) singular

Best explanation: SINCE, M is a lower triangular matrix with diagonal ELEMENTS zero, then we add I and it will RESULT in a lower triangular matrix with all diagonal entries 1. Thus, all EIGENVALUES M+I are non zero (eigenvalues of triangular matrix is the diagonal elements). So, determinant will never be zero. Hence, the matrix can have inverse property.

The CORRECT option is (b) Y TRANSPOSE is nilpotent

The explanation: Suppose, we have a matrix

\(a=\begin{bmatrix}

1 & 0\\

2 & 1\\

\end{bmatrix} \)

 then Y^2 will not resulting in Y, hence it is not IDEMPOTENT, Y^2 is not 0 and so it is not nilpotent. But, as Y is a square matrix, by the property inverse will exist in this case.

Right OPTION is (a) -(X+4)

For EXPLANATION: Let, Identity element I, x*I = I*x = x ⇒ x = x + I – 2 ⇒ I = 2. INVERSE of x is x^-1

Now, x*x^-1 = I

⇒ x + x^-1 – 2 = 2

⇒ x^-1 = -(x+4).

Correct answer is (a) 3

Explanation: A symmetric group on a SET of three elements is SAID to be the group of all permutations of a three-element set. It is a DIHEDRAL group of order six having degree three.

The correct choice is (b) a^n = 0, for some positive integer n

Explanation: Since a is nilpotent in a commutative ring R, we have an=0 for some positive integer n. since R is commutative, for any m∈R, we have (am)n=anmn=0. Then we have the FOLLOWING equality: (1−am)(1+(am)+(am)2+⋯+(am)n−1)=1. Hence, 1−am is a UNIT in R.

The correct option is (a) solvable

The best I can EXPLAIN: The prime factorization of 20 is 20=2⋅5. Let n5 be the number of 5-Sylow subgroups of G. By Sylow’s theorem, we have,n5≡1(mod 5)and n5|4. THUS, we have n5=1. Let P be the unique 5-Sylow subgroup of G. The subgroup P is normal in G as it is the unique 5-Sylow subgroup. Then consider the subnormal series G▹P▹{e}, where e is the identity element of G. Then the factor GROUPS G/P, P/{e} have order 4 and 5 respectively. HENCE these are cyclic groups(in particular ABELIAN). Hence, the group G of order 20 has a subnormal series whose factor groups are abelian groups, and thus G is a solvable group.

Correct option is (d) commutative

To EXPLAIN: LET R be a ring with unit 1. Suppose that the ORDER of R is |R|=p2 for some prime NUMBER p. Then it has been proven that R is a commutative ring.

The correct option is (a) infinite

Explanation: Let p be the order of g (hence the order of G). As a contradiction, ASSUME that p=ab is a composite NUMBER with integers a > 1, b > 1. Then (ga) is a proper normal subgroup of G. This is a contradiction since G is SIMPLE. Thus, p MUST be a PRIME number.

Therefore, the order of G is a prime number.

The correct answer is (C) ODD

For explanation I would say: LET g≠e be an element in group F such that g7=e. As 7 is a prime number, this yields that the ORDER of g is 7. Consider, the subgroup (g) is generated by g. As the order of g is 7, the order of the subgroup (g)is 7. Hence, the order MUST be odd.

The CORRECT option is (a) 144

For explanation I WOULD say: The NUMBER of generators of a cyclic group of order N is equal to the number of integers between 1 and n that are relatively prime to n.Namely, the number of generators is equal to ϕ(n), where ϕ is the Euler totient function. We know that G is a cyclic group of order 219. Hence, the number of generators of G is ϕ(219) = ϕ(3)ϕ(73) = 3⋅73 = 144.

The correct option is (d) a cyclic group

Easy explanation: The prime factorization 219=3⋅73. By the definition of Sylow’s theorem, determine the number np of Sylow p-group for p=3,73. np≡1(MOD p) and np divides n/p. THUS, n3 could be 1, 4, 7, 10, 13,… and n3 NEEDS to divide 219/3=73. Hence the only possible value for n3 is n3=1. So there is a unique Sylow 3-subgroup P3 of G. By Sylow’s theorem, the unique Sylow 3-subgroup must be a normal subgroup of G. Similarly, n73=1, 74,… and n73 must divide 219/73=3 and hence we must have n73=1. Thus, G has a unique normal Sylow 73-subgroup P73.

Correct answer is (b) abelian group

Explanation: LET C be a cyclic group with a generator g∈C. Namely, we have G={g.Let x and y be arbitrary elements in C. Then, there exists N, m∈Z such that x=gn and y=gm. It FOLLOWS that x*y = gn*gm = gn+m = gm*gn = yx. Hence, we find that xy=yx for any x,y belongs to G.Thus, G is an abelian group.

Correct choice is (c) Composite

To elaborate: Suppose that any finite group of order less than n has a composition series. Let G be a finite group of order n. If G is simple, then G⊳{e}, where e is the identity element of G and HENCE, it is a composition series. HOWEVER, any INFINITE cyclic group does not have a composite series.

Right choice is (c) 2

To explain: The number of Abelian groups of order P^m (let, P is prime) is the number of partitions of m. Here order is 8 i.e. 2^3 and so PARTITION of 3 are {1, 1} and {3, 0}. So number of different abelian groups are 2.

Right choice is (d) transversal

The best I can explain: A coset representative is a representative in the EQUIVALENCE CLASS. In all COSETS, a SET of the representative is always transversal.

The correct OPTION is (c) The set of all non-singular matrices forms a GROUP under multiplication

Easiest explanation: Since multiplication of two NEGATIVE rational numbers gives a positive number. Hence, closure property is not satisfied. Singular matrices do not form a group under multiplication. Matrices have to be non-singular (determinant !=0) for the INVERSE to exist. Hence the set of all non-singular matrices forms a group under multiplication is a true option.

Correct OPTION is (a) abelian group

The EXPLANATION: An example of a COSET is associated with the theory of vector spaces. The elements (VECTORS) form an abelian group under the vector addition in a vector space. Subspaces of a vector space are subgroups of this group.

Correct answer is (b) the ORDER of the SUBGROUP divides the order of the finite GROUP

The explanation is: Lagrange’s theorem SATISFIES that the order of the subgroup divides the order of the finite group.

The CORRECT OPTION is (d) H is subgroup of an abelian GROUP

For explanation: If h is the subgroup of an abelian group G, then the set of left cosets of H in G is to be set of right cosets i.e, a * H = H * a. Hence, subgroup is CALLED the normal subgroup.

The CORRECT option is (C) isomorphism

Best explanation: Two GROUPS M and K are isomorphic (M ~= K) if and only if there EXISTS an isomorphism between them. An isomorphism f:M -> K between two groups M and K is a mapping which satisfies two conditions: 1) f is a bijection and 2) for every x,y belongs to M, we have f(x*My) = f(x) * Kf(y).

Correct answer is (b) left

The best explanation: Let (H, *) be the SEMIGROUP of the GROUP (G, *). Let a BELONGS to G. (a * H) is the set of a left coset of H in G and (H * a) be the set of a right coset of H in G.

Right option is (d) an invariant EQUIPPED with conjugation by the elements of original group

To ELABORATE: A normal subgroup is a subgroup that is invariant under conjugation by any element of the original group that is, K is normal if and only if gKg-1=K for any G BELONGS to G Equivalently, a subgroup K of G is normal if and only if gK=Kg for any g belongs to G.Normal subgroups are useful in constructing quotient groups and in analyzing HOMOMORPHISMS.

The correct option is (a) a subgroup complex NUMBERS having magnitude 1 of the GROUP of NONZERO complex elements

Best explanation: The SET of complex numbers with magnitude 1 is a subgroup of the nonzero complex numbers associated with multiplication. It is CALLED the circle group as its elements form the unit circle.

The CORRECT choice is (B) subgroup

For explanation: The subgroup property is intersection closed. An ARBITRARY (nonempty) intersection of SUBGROUPS with this property, ALSO attains the similar property.